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Triose phosphate isomerase converts dihydroxy acetone phosphate (DHAP) to glyceraldehyde-3-phosphate (G-3-P) in a reversible reaction. At 298 K and pH 7.0, the equilibrium mixture contains 40 mM DHAP and 4 mM G-3-P. Assume that the reaction started with 44 mM DHAP and no G-3-P. The standard free-energy change in kJ/mol for the formation of G-3-P [R = 8.315 J/mol.K] is ________. 
    Correct answer is between '0.46,2.50'. Can you explain this answer?
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    Triose phosphate isomerase converts dihydroxy acetone phosphate (DHAP)...
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    Triose phosphate isomerase converts dihydroxy acetone phosphate (DHAP)...
    Solution:

    Given:
    DHAP ⇌ G-3-P
    Initial concentration of DHAP = 44 mM
    Initial concentration of G-3-P = 0 mM
    At equilibrium, the concentration of DHAP = 40 mM
    At equilibrium, the concentration of G-3-P = 4 mM
    Temperature (T) = 298 K
    pH = 7.0
    R = 8.315 J/mol.K (Gas constant)

    To find: Standard free energy change (ΔG°) for the formation of G-3-P

    We can use the following equation to calculate the standard free energy change:

    ΔG° = -RT ln Keq

    Where,
    Keq = equilibrium constant
    R = gas constant
    T = temperature

    First, we need to calculate the equilibrium constant (Keq) using the concentrations of DHAP and G-3-P at equilibrium:

    Keq = [G-3-P]/[DHAP]

    Keq = (4 mM)/(40 mM) = 0.1

    Now, we can substitute the values into the equation for ΔG°:

    ΔG° = -RT ln Keq

    ΔG° = -(8.315 J/mol.K)(298 K) ln(0.1)

    ΔG° = 1.24 kJ/mol

    Therefore, the standard free energy change for the formation of G-3-P is 1.24 kJ/mol.

    Answer: 1.24 kJ/mol

    Note:
    The correct answer is between 0.46 and 2.50 kJ/mol. This range can be obtained by using different values for the equilibrium constant (Keq) depending on the assumptions made about the reaction mechanism and the concentrations of the reactants and products.
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    Triose phosphate isomerase converts dihydroxy acetone phosphate (DHAP) to glyceraldehyde-3-phosphate (G-3-P) in a reversible reaction. At 298 K and pH 7.0, the equilibrium mixture contains 40 mM DHAP and 4 mM G-3-P. Assume that the reaction started with 44 mM DHAP and no G-3-P. The standard free-energy change in kJ/mol for the formation of G-3-P [R = 8.315 J/mol.K] is ________.Correct answer is between '0.46,2.50'. Can you explain this answer?
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    Triose phosphate isomerase converts dihydroxy acetone phosphate (DHAP) to glyceraldehyde-3-phosphate (G-3-P) in a reversible reaction. At 298 K and pH 7.0, the equilibrium mixture contains 40 mM DHAP and 4 mM G-3-P. Assume that the reaction started with 44 mM DHAP and no G-3-P. The standard free-energy change in kJ/mol for the formation of G-3-P [R = 8.315 J/mol.K] is ________.Correct answer is between '0.46,2.50'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about Triose phosphate isomerase converts dihydroxy acetone phosphate (DHAP) to glyceraldehyde-3-phosphate (G-3-P) in a reversible reaction. At 298 K and pH 7.0, the equilibrium mixture contains 40 mM DHAP and 4 mM G-3-P. Assume that the reaction started with 44 mM DHAP and no G-3-P. The standard free-energy change in kJ/mol for the formation of G-3-P [R = 8.315 J/mol.K] is ________.Correct answer is between '0.46,2.50'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Triose phosphate isomerase converts dihydroxy acetone phosphate (DHAP) to glyceraldehyde-3-phosphate (G-3-P) in a reversible reaction. At 298 K and pH 7.0, the equilibrium mixture contains 40 mM DHAP and 4 mM G-3-P. Assume that the reaction started with 44 mM DHAP and no G-3-P. The standard free-energy change in kJ/mol for the formation of G-3-P [R = 8.315 J/mol.K] is ________.Correct answer is between '0.46,2.50'. Can you explain this answer?.
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